There are balls of 3 colours in a bag: 3 red, 4 green, 5 blue. Randomly perform 5 draws. In each draw, retrieve 3 balls from the bag at the same time . And place the balls back and perform next draw. What is the probability of getting 3 balls of same color in any 3 draws (means excluding 1,2,4,5 draws of the same color)?

  • $\begingroup$ Not sure this is clear. It appears that you are asking to get three balls of the same colors at least three times (out of the five tries). Is that correct? I assume the balls are replaced each time, yes? Where does your formula come from? What is the meaning of all the factors of $\binom 53$? $\endgroup$ – lulu Jul 19 '18 at 22:08
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    $\begingroup$ Please edit your post for clarity. If my interpretation is correct (of which I am not at all sure) then figure out the probability $p$ that a single drawing is monochromatic. Once you've done that, the answer is a simple binomial calculation. $\endgroup$ – lulu Jul 19 '18 at 22:11
  • $\begingroup$ I've edited the question and clarified. thanks. $\endgroup$ – techie11 Jul 19 '18 at 22:13
  • $\begingroup$ My question about replacement concerned replacement after each draw (of three). As there are only $12$ balls in total, if you don't replace them, you will run out. So I assume that you don't replace while drawing the three but then you replace the three for the next draw, yes? Assuming that, then the method I sketched should work. Compute $p$, the probability that a single draw is one color, then use the binomial distribution to conclude (since draws of three elements are independent of each other). $\endgroup$ – lulu Jul 19 '18 at 22:16
  • $\begingroup$ I did furthur editing to clarify. $\endgroup$ – techie11 Jul 19 '18 at 22:20

Stipulation: What we want, I'm assuming, is the probability that out of $5$ three-ball draws, exactly $3$ draws have all three balls with the same color (either red or green or blue).

First, find the probability that any given draw consists of just one color.

\begin{align} P(\text{all one color}) & = P(\text{all red}) + P(\text{all green}) + P(\text{all blue}) \\ & = \frac{\binom{3}{3}}{\binom{12}{3}} + \frac{\binom{4}{3}}{\binom{12}{3}} + \frac{\binom{5}{3}}{\binom{12}{3}} \end{align}

Let this quantity be denoted $q$. Then you can just use the binomial theorem to determine the probability that exactly three of these three-ball draws end up with all one color:

$$ P(\text{exactly $3$ of the $5$ three-ball draws are all one color}) = \binom{5}{3} q^3 (1-q)^2 $$

  • $\begingroup$ Thank you. Brian. What if a second question is asked: how many number of outcomes this event contains? $\endgroup$ – techie11 Jul 19 '18 at 22:39
  • $\begingroup$ @techie11: Are you asking that because of the Terence Tao question? $\endgroup$ – Brian Tung Jul 19 '18 at 22:46
  • $\begingroup$ not actually. I just trying to understand the basic idea about counting :-) $\endgroup$ – techie11 Jul 19 '18 at 22:50
  • $\begingroup$ @techie11 Clearly there are $\binom 53\, {\left(\binom 33+\binom 43+\binom 53\right)}^3\,{\left(\binom{12}3-(\binom 33+\binom 43+\binom 53)\right)}^2$ favoured outcomes out of $\binom{12}3^5$ total. $\endgroup$ – Graham Kemp Jul 20 '18 at 1:42
  • $\begingroup$ @Graham Kemp that is obvious. thanks. $\endgroup$ – techie11 Jul 20 '18 at 19:26

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